Question
Download Solution PDFWhich one of the following is not a basic solution of system of linear equation
x1 + 2x2 + x3 = 4
2x1 + x2 + 5x3 = 5
Answer (Detailed Solution Below)
Option 1 : x1 = -1, x2 = 2, x3 = 1
Detailed Solution
Download Solution PDFThe correct answer is : option 1
Key Points
Concept of Basic Solution:
- A basic solution to a system of linear equations in matrix form (Ax = b) is obtained by setting (n - m) variables to zero, where:
- n = total number of variables
- m = number of equations
- Here, we have 2 equations (m = 2) and 3 variables (x₁, x₂, x₃), so:
- We must set (3 - 2) = 1 variable to zero to get a basic solution.
- If all variables are non-zero, it is not a basic solution.
Now, let’s analyze each option:
Option 1) x₁ = -1, x₂ = 2, x₃ = 1
- All variables are non-zero → Not a valid basic solution.
- Check equation 1: -1 + 2×2 + 1 = -1 + 4 + 1 = 4 ✔
- Check equation 2: 2×(-1) + 2 + 5×1 = -2 + 2 + 5 = 5 ✔
- This is NOT a basic solution because no variable is zero.
Option 2) x₁ = 2, x₂ = 1, x₃ = 0
- x₃ = 0 (one variable is zero) → could be a basic solution.
- Equation 1: 2 + 2×1 + 0 = 4 ✔
- Equation 2: 2×2 + 1 + 5×0 = 4 + 1 = 5 ✔
- Valid basic solution.
Option 3) x₁ = 5, x₂ = 0, x₃ = -1
- x₂ = 0 → One variable zero → Candidate for basic solution.
- Equation 1: 5 + 0 + (-1) = 4 ✔
- Equation 2: 2×5 + 0 + 5×(-1) = 10 - 5 = 5 ✔
- Valid basic solution.
Option 4) x₂ = 5/3, x₃ = 2/3
- Need to find x₁:
- Equation 1: x₁ + 2×(5/3) + (2/3) = 4 → x₁ + 10/3 + 2/3 = 4 → x₁ = 4 - 12/3 = 0
- x₁ = 0 → One variable is zero → Possible basic solution
- Equation 2: 2×0 + 5/3 + 5×(2/3) = 5/3 + 10/3 = 15/3 = 5 ✔
- Valid basic solution.
✅ Final Answer:
Option 1) x₁ = -1, x₂ = 2, x₃ = 1 is NOT a basic solution because all variables are non-zero.